find the directional derivative of f x y z

Scholarship Objectives

• 4.6.1 Determine the directing derivative in a presumption direction for a function of two variables.
• 4.6.2 Determine the gradient vector of a given real-valued function.
• 4.6.3 Explain the implication of the gradient transmitter with regard to direction of change on a surface.
• 4.6.4 Economic consumption the gradient to find the tangent to a level curve of a given function.
• 4.6.5 Calculate directional derivatives and gradients in three dimensions.

In Partial Derivatives we introduced the overtone derivative instrument. A function $z=f\left(x,y\right)$ has deuce partial derivatives: $\partial z\text{/}\partial x$ and $\partial z\text{/}\partial y.$ These derivatives correspond to each of the independent variables and privy be interpreted as instantaneous rates of exchange (that is, as slopes of a tangent line). For example, $\partial z\text{/}\partial x$ represents the gradient of a tangent melodic line cursory through a given point on the surface defined by $z=f\left(x,y\right),$ forward the tan line is parallel to the x-Axis. Similarly, $\partial z\text{/}\partial y$ represents the slope of the tan note parallel to the $y\text{-axis.}$ Now we consider the hypothesis of a tan line collimate to neither bloc.

Directive Derivatives

We start with the graph of a superficial distinct by the equation $z=f\left(x,y\right).$ Given a point $\left(a,b\right)$ in the domain of $f,$ we choose a direction to travel from that point. We measure the direction using an angle $\theta ,$ which is measured counterclockwise in the x, y-carpenter's plane, starting at nada from the positive x-axis (Figure 4.39). The distance we travel is $h$ and the direction we move out is given by the unit of measurement vector $u=\left(\text{romaine lettuce}\theta \right)i+\left(\text{goof}\theta \right)j.$ Therefore, the z-equal of the forward point on the graph is given by $z=f\left(a+h\text{cos}\theta ,b+h\text{sin}\theta \right).$

Figure 4.39 Determination the directional derived at a charge connected the graph of $z=f\left(x,y\right).$ The slope of the black pointer on the graph indicates the value of the directional derivative at that point.

We can calculate the slope of the sec line away dividing the divergence in $z\text{-values}$ past the duration of the line segment conjunctive the two points in the domain. The distance of the line segment is $h.$ Therefore, the gradient of the secant line is

$$m_{\text{sec}}=\frac{f\left(a+h\text{cosine}\theta ,b+h\text{sin}\theta \right)-f\left(a,b\right)}{h}.$$

To find the slope of the tan line in the equivalent direction, we take the fix as $h$ approaches zero.

Definition

Suppose $z=f\left(x,y\right)$ is a function of two variables with a domain of $D.$ Let $\left(a,b\right)\in D$ and define $\text{u}=\text{romaine lettuce}\theta i+\text{sin}\theta j.$ Then the guiding derivative of $f$ in the direction of $u$ is apt by

$$D_{u}f\left(a,b\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(a+h\text{cos}\theta ,b+h\text{hell}\theta \right)-f\left(a,b\right)}{h},$$

(4.36)

provided the limit exists.

Equation 4.36 provides a formal definition of the directional derivative instrument that butt be misused in many cases to calculate a guiding derivative.

Example 4.31

Finding a Directing Derivative instrument from the Definition

Let $\theta =\text{arc cosine}\left(3\text{/}5\right).$ Get the directional derivative $D_{u}f\left(x,y\right)$ of $f\left(x,y\right)=x^2-xy+3y^2$ in the direction of $\text{u}=\left(\text{romaine lettuce}\theta \right)i+\left(\text{sin}\theta \right)j.$ What is $D_{u}f\left(\mathrm{-1},2\right)?$

Another approach to calculating a directional derivative involves partial derivatives, as defined in the following theorem.

Theorem 4.12

Directive Derivative of a Social function of Two Variables

Let $z=f\left(x,y\right)$ follow a function of two variables $x\text{and}y,$ and assume that $f_x$ and $f_y$ exist. Then the directional derivative of $f$ in the direction of $\text{u}=\text{romaine lettuce}\theta \text{i}+\text{sin}\theta \text{j}$ is acknowledged by

$$D_{u}f\left(x,y\right)=f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{goof}\theta .$$

(4.37)

Proof

Par 4.36 states that the position derivative of f in the direction of $u=\text{cos}\theta i+\text{sin}\theta j$ is relinquished aside

$$D_{u}f\left(a,b\right)=\underset{t\to 0}{\text{lim}}\frac{f\left(a+t\text{romaine lettuce}\theta ,b+t\text{sin}\theta \right)-f\left(a,b\right)}{t}.$$

Let $x=a+t\text{cos}\theta$ and $y=b+t\text{sin}\theta ,$ and delineate $g\left(t\right)=f\left(x,y\right).$ Since $f_x$ and $f_y$ both exist, and therefore $f$ is differentiable, we can use the mountain chain rule out for functions of two variables to calculate $g^{\prime }\left(t\right)\text{:}$

$$\begin{array}{cc}\hfill g^{\prime }\left(t\right)& =\frac{\partial \mathit{\text{f}}}{\partial \mathit{\text{x}}}\frac{dx}{dt}+\frac{\partial \mathit{\text{f}}}{\partial \mathit{\text{y}}}\frac{dy}{dt}\hfill \\ & =f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{sin}\theta .\hfill \end{array}$$

If $t=0,$ past $x=x_0(=a)$ and $y=y_0(=b),$ so

$$g^{\prime }\left(0\right)=f_x\left(x_0,y_0\right)\text{cos}\theta +f_y\left(x_0,y_0\right)\text{sin}\theta .$$

By the definition of $g^{\prime }\left(t\right),$ IT is likewise dependable that

$$\begin{array}{cc}\hfill g^{\prime }\left(0\right)& =\underset{t\to 0}{\text{lim}}\frac{g\left(t\right)-g\left(0\right)}{t}\hfill \\ & =\underset{t\to 0}{\text{lim}}\frac{f\left(x_0+t\text{cos}\theta ,y_0+t\text{sin}\theta \right)-f\left(x_0,y_0\right)}{t}.\hfill \end{array}$$

Therefore, $D_{u}f\left(x_0,y_0\right)=f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{sin}\theta .$

□

Example 4.32

Finding a Directional Derivative: Alternative Method

Let $\theta =\text{arccos}\left(3\text{/}5\right).$ Find the directional derivative $D_{u}f\left(x,y\right)$ of $f\left(x,y\right)=x^2-xy+3y^2$ in the direction of $u=\left(\text{cos lettuce}\theta \right)i+\left(\text{sin}\theta \right)j.$ What is $D_{u}f\left(\mathrm{-1},2\right)?$

Checkpoint 4.28

Find the directional derivative instrument $D_{u}f\left(x,y\right)$ of $f\left(x,y\right)=3x^{2}y-4x{y}^3+3y^2-4x$ in the direction of $u=\left(\text{cos}\frac{\pi }{3}\right)i+\left(\text{sin}\frac{\pi }{3}\right)j$ using Equation 4.37. What is $D_{u}f\left(3,4\right)?$

If the vector that is tending for the commission of the derivative is not a unit transmitter, then it is only necessity to split by the average of the vector. For example, if we wished to find the directional derivative of the function in Illustration 4.32 in the direction of the vector $\langle \mathrm{-5},12\rangle ,$ we would first part by its order of magnitude to get $u.$ This gives us $u=\langle \text{−}\left(5\text{/}13\right),12\text{/}13\rangle .$ Then

$$\begin{array}{cc}\hfill D_{u}f\left(x,y\right)& =\nabla f\left(x,y\right)·u\hfill \\ & =-\frac{5}{13}\left(2x-y\right)+\frac{12}{13}\left(\text{−}x+6y\right)\hfill \\ & =-\frac{22}{13}x+\frac{17}{13}y.\hfill \end{array}$$

Slope

The helpful side of Equation 4.37 is isochronous to $f_x\left(x,y\right)\text{cos}\theta +f_y\left(x,y\right)\text{sin}\theta ,$ which nates be written as the dot product of cardinal vectors. Define the for the first time vector as $\nabla f\left(x,y\right)=f_x\left(x,y\right)\text{i}+f_y\left(x,y\right)\text{j}$ and the second vector as $u=\left(\text{cos}\theta \right)i+\left(\text{sin}\theta \right)j.$ Then the right-hand side of the equation can be written as the dot product of these two vectors:

$$D_{u}f\left(x,y\right)=\nabla f\left(x,y\right)·u.$$

(4.38)

The first transmitter in Equation 4.38 has a special list: the gradient of the function $f.$ The symbolization $\nabla$ is called nabla and the transmitter $\nabla f$ is show $\text{"del}f\text{."}$

Definition

Have $z=f\left(x,y\right)$ be a function of $x\text{and}y$ such that $f_x$ and $f_y$ subsist. The transmitter $\nabla f\left(x,y\right)$ is called the gradient of $f$ and is defined American Samoa

$$\nabla f\left(x,y\right)=f_x\left(x,y\right)i+f_y\left(x,y\right)j.$$

(4.39)

The vector $\nabla f\left(x,y\right)$ is also written as $\text{"grade}f\text{."}$

Example 4.33

Finding Gradients

Find the gradient $\nabla f\left(x,y\right)$ of apiece of the following functions:

1. $f\left(x,y\right)=x^2-xy+3y^2$
2. $f\left(x,y\right)=\text{sin}3x\text{cos}3y$

Checkpoint 4.29

Regain the gradient $\nabla f\left(x,y\right)$ of $f(x,y)=\left(x^2-3y^2\right)\text{/}\left(2x+y\right).$

The gradient has any important properties. We have already seen combined formula that uses the gradient: the formula for the leading derivative. Recall from The DoT Product that if the angle between two vectors $a$ and $b$ is $\phi ,$ then $a·\text{b}=\Vert a\Vert \Vert \text{b}\Vert \text{cos}\phi .$ Therefore, if the angle between $\nabla f\left(x_0,y_0\right)$ and $u=\left(\text{romaine lettuce}\theta \right)i+\left(\text{sin}\theta \right)j$ is $\phi ,$ we make

$$D_{u}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·u=\Vert \nabla f\left(x_0,y_0\right)\Vert \Vert u\Vert \text{cos}\phi =\Vert \nabla f\left(x_0,y_0\right)\Vert \text{cos}\phi .$$

The $\Vert u\Vert$ disappears because $u$ is a unit vector. Therefore, the directing derivative is equal to the order of magnitude of the slope evaluated at $\left(x_0,y_0\right)$ multiplied away $\text{cos}\phi .$ Recall that $\text{cos}\phi$ ranges from $\mathrm{-1}$ to $1.$ If $\phi =0,$ and then $\text{cos}\phi =1$ and $\nabla f\left(x_0,y_0\right)$ and $u$ both point in the Same direction. If $\phi =\pi ,$ then $\text{cos}\phi =\mathrm{-1}$ and $\nabla f\left(x_0,y_0\right)$ and $u$ point in opposite directions. In the first case, the note value of $D_{\text{u}}f\left(x_0,y_0\right)$ is maximized; in the second case, the value of $D_{\text{u}}f\left(x_0,y_0\right)$ is minimized. If $\nabla f\left(x_0,y_0\right)=0,$ then $D_{u}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·u=0$ for whatever vector $u.$ These three cases are outlined in the favorable theorem.

Theorem 4.13

Properties of the Slope

Theorise the social function $z=f\left(x,y\right)$ is differentiable at $\left(x_0,y_0\right)$ (Figure 4.41).

1. If $\nabla f\left(x_0,y_0\right)=0,$ then $D_{u}f\left(x_0,y_0\right)=0$ for whatever unit vector $u.$
2. If $\nabla f\left(x_0,y_0\right)\ne 0,$ then $D_{u}f\left(x_0,y_0\right)$ is maximized when $u$ points in the same direction as $\nabla f\left(x_0,y_0\right).$ The maximum note value of $D_{u}f\left(x_0,y_0\right)$ is $\Vert \nabla f\left(x_0,y_0\right)\Vert \text{.}$
3. If $\nabla f\left(x_0,y_0\right)\ne 0,$ past $D_{u}f\left(x_0,y_0\right)$ is reduced when $u$ points in the opposite counsel from $\nabla f\left(x_0,y_0\right).$ The nominal value of $D_{u}f\left(x_0,y_0\right)$ is $\text{−}\Vert \nabla f\left(x_0,y_0\right)\Vert \text{.}$

Figure 4.41 The gradient indicates the maximum and minimum values of the social control derivative at a point.

Exercise 4.34

Finding a Maximum Directional Differential

Find the direction for which the directional differential of $f\left(x,y\right)=3x^2-4xy+2y^2$ at $\left(\mathrm{-2},3\right)$ is a upper limit. What is the uttermost value?

Checkpoint 4.30

Bump the direction for which the directional derivative of $g\left(x,y\right)=4x-xy+2y^2$ at $\left(\mathrm{-2},3\right)$ is a maximum. What is the maximum value?

Material body 4.43 shows a percentage of the graph of the use $f\left(x,y\right)=3+\text{sin}x\text{sin}y.$ Given a point $\left(a,b\right)$ in the domain of $f,$ the maximal apprais of the gradient at that point is given by $\Vert \nabla f\left(a,b\right)\Vert .$ This would equal the value of greatest ascent if the surface represented a geographics map. If we went in the opposite direction, it would be the rate of greatest descent.

Human body 4.43 A typical surface in ${ℝ}^3.$ Given a point on the rise up, the directional derivative can be calculated victimization the slope.

When victimization a topographical map, the steepest slope is always in the charge where the contour line lines are nearest in concert (see Figure 4.44). This is correspondent to the contour map of a function, forward the level curves are obtained for equally spaced values end-to-end the range of that subprogram.

Figure 4.44 Contour map for the function $f\left(x,y\right)=x^2-y^2$ victimisation level values between $\mathrm{-5}$ and $5.$

Gradients and Level Curves

Recall that if a curve is defined parametrically aside the function pair $\left(x\left(t\right),y\left(t\right)\right),$ then the vector $x^{\prime }\left(t\right)i+y^{\prime }\left(t\right)j$ is tangent to the curve for every evaluate of $t$ in the domain. Now let's arrogate $z=f\left(x,y\right)$ is a differentiable function of $x\text{and}y,$ and $\left(x_0,y_0\right)$ is in its domain. Have's conjecture encourage that $x_0=x\left(t_0\right)$ and $y_0=y\left(t_0\right)$ for extraordinary value of $t,$ and think the level curve $f\left(x,y\right)=k.$ Define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)$ and calculate $g^{\prime }\left(t\right)$ on the level curve. By the chain Formula,

$$g^{\prime }\left(t\right)=f_x\left(x\left(t\right),y\left(t\right)\right)x^{\prime }\left(t\right)+f_y\left(x\left(t\right),y\left(t\right)\right)y^{\prime }\left(t\right).$$

But $g^{\prime }\left(t\right)=0$ because $g\left(t\right)=k$ for all $t.$ Therefore, on one hand,

$$f_x\left(x\left(t\right),y\left(t\right)\right)x^{\prime }\left(t\right)+f_y\left(x\left(t\right),y\left(t\right)\right)y^{\prime }\left(t\right)=0;$$

then again,

$$f_x\left(x\left(t\right),y\left(t\right)\right)x^{\prime }\left(t\right)+f_y\left(x\left(t\right),y\left(t\right)\right)y^{\prime }\left(t\right)=\nabla f\left(x,y\right)·\langle x^{\prime }\left(t\right),y^{\prime }\left(t\right)\rangle \text{.}$$

Therefore,

$$\nabla f\left(x,y\right)·\langle x^{\prime }\left(t\right),y^{\prime }\left(t\right)\rangle =0.$$

So, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.

Theorem 4.14

Gradient Is Normal to the Level Curve

Suppose the function $z=f\left(x,y\right)$ has continuous first-fiat partial derivatives in an open disk centered at a manoeuvre $\left(x_0,y_0\right).$ If $\nabla f\left(x_0,y_0\right)\ne 0,$ then $\nabla f\left(x_0,y_0\right)$ is pattern to the level curve of $f$ at $\left(x_0,y_0\right).$

We can use this theorem to get tan and normal vectors to level curves of a function.

Deterrent example 4.35

Finding Tangents to Point Curves

For the function $f\left(x,y\right)=2x^2-3xy+8y^2+2x-4y+4,$ find a tan transmitter to the level curve at point $\left(\mathrm{-2},1\right).$ Graphical record the horizontal surface curve corresponding to $f\left(x,y\right)=18$ and draw in $\nabla f\left(\mathrm{-2},1\right)$ and a tangent vector.

Checkpoint 4.31

For the function $f\left(x,y\right)=x^2-2xy+5y^2+3x-2y+4,$ find the tan to the pull dow curve at point $\left(1,1\right).$ Draw the graphical record of the level curve corresponding to $f\left(x,y\right)=8$ and thread $\nabla f\left(1,1\right)$ and a tangent vector.

Three-Dimensional Gradients and Directional Derivatives

The definition of a gradient fanny glucinium extended to functions of more than cardinal variables.

Definition

Let $w=f\left(x,y,z\right)$ be a function of three variables such that $f_x,f_y,\text{and}{f}_z$ subsist. The transmitter $\nabla f\left(x,y,z\right)$ is called the gradient of $f$ and is defined as

$$\nabla f\left(x,y,z\right)=f_x\left(x,y,z\right)i+f_y\left(x,y,z\right)j+f_z\left(x,y,z\right)k.$$

(4.40)

$\nabla f\left(x,y,z\right)$ give the axe also comprise written as $\text{grad}f\left(x,y,z\right).$

Calculating the gradient of a function in three variables is very confusable to calculating the gradient of a serve in ii variables. First, we calculate the partial derivatives $f_x,f_y,$ and $f_z,$ and then we use Equation 4.40.

Example 4.36

Determination Gradients in Three Dimensions

Find the gradient $\nabla f\left(x,y,z\right)$ of each of the favourable functions:

1. $f\left(x,y,z\right)=5x^2-2xy+y^2-4yz+z^2+3xz$
2. $f\left(x,y,z\right)=e^{\mathrm{-2}z}\text{trespass}2x\text{cos lettuce}2y$

Checkpoint 4.32

Find the gradient $\nabla f\left(x,y,z\right)$ of $f(x,y,z)=\frac{x^2-3y^2+z^2}{2x+y-4z}.$

The directional derivative can likewise be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needful. This vector is a building block vector, and the components of the unit vector are called directional cosines . Given a 3-dimensional building block vector $u$ in standard variety (i.e., the initial point is at the pedigree), this vector forms three different angles with the cocksure $x-,y-,$ and z-axes. Lashkar-e-Tayyiba's call these angles $\alpha ,\beta ,$ and $\gamma .$ Past the directional cosines are given by $\text{cosine}\alpha ,\text{cos}\beta ,$ and $\text{cos}\gamma .$ These are the components of the social unit transmitter $u;$ since $u$ is a unit vector, it is true that ${\text{cos lettuce}}^2\alpha +{\text{cos}}^2\beta +{\text{cos lettuce}}^2\gamma =1.$

Definition

Conjecture $w=f\left(x,y,z\right)$ is a function of three variables with a domain of $D.$ Allow $\left(x_0,y_0,z_0\right)\in D$ and Lashkar-e-Tayyiba $\text{u}=\text{cos}\alpha i+\text{cos}\beta j+\text{cos}\gamma k$ be a unit vector. Then, the directional derivative of $f$ in the direction of $u$ is apt by

$$D_{u}f\left(x_0,y_0,z_0\right)=\underset{t\to 0}{\text{lim}}\frac{f\left(x_0+t\text{cosine}\alpha ,y_0+t\text{cos}\beta ,z_0+t\text{cos}\gamma \right)-f\left(x_0,y_0,z_0\right)}{t},$$

(4.41)

provided the limit exists.

We can calculate the social control derivative of a function of trey variables by using the gradient, leading to a formula that is analogous to Equation 4.38.

Theorem 4.15

Directional Differential coefficient of a Function of Three Variables

Let $f\left(x,y,z\right)$ personify a differentiable function of three variables and let $u=\text{cos lettuce}\alpha i+\text{cos lettuce}\beta j+\text{cos}\gamma k$ follow a unit vector. And then, the directional derivative of $f$ in the focusing of $u$ is given by

$$\begin{array}{cc}\hfill D_{u}f\left(x,y,z\right)& =\nabla f\left(x,y,z\right)·u\hfill \\ & =f_x\left(x,y,z\right)\text{cos}\alpha +f_y\left(x,y,z\right)\text{cos}\beta +f_z\left(x,y,z\right)\text{cos}\gamma .\hfill \end{array}$$

(4.42)

The three angles $\alpha ,\beta ,\text{and}\gamma$ determine the unit transmitter $u.$ In practice, we can use an arbitrary (nonunit) transmitter, then carve up by its magnitude to incur a unit vector in the desirable direction.

Deterrent example 4.37

Determination a Leading Derivative in Three Dimensions

Calculate $D_{u}f\left(1,\mathrm{-2},3\right)$ in the direction of $\text{v}=\text{−}i+2j+2k$ for the routine

$$f\left(x,y,z\right)=5x^2-2xy+y^2-4yz+z^2+3xz.$$

Checkpoint 4.33

Count on $D_{u}f\left(x,y,z\right)$ and $D_{u}f\left(0,\mathrm{-2},5\right)$ in the direction of $\text{v}=\mathrm{-3}i+12j-4k$ for the function $f\left(x,y,z\right)=3x^2+xy-2y^2+4yz-z^2+2xz.$

Section 4.6 Exercises

For the succeeding exercises, find the directional derivative using the limit definition only.

260 .

$f\left(x,y\right)=5-2x^2-\frac{1}{2}{y}^2$ at point $P\left(3,4\right)$ in the direction of $\text{u}=\left(\text{romaine}\frac{\pi }{4}\right)i+\left(\text{sin}\frac{\pi }{4}\right)j$

261 .

$f\left(x,y\right)=y^2\text{cos}\left(2x\right)$ at taper $P\left(\frac{\pi }{3},2\right)$ in the direction of $\text{u}=\left(\text{cos}\frac{\pi }{4}\right)i+\left(\text{wickedness}\frac{\pi }{4}\right)j$

262 .

Find the directional derivative of $f\left(x,y\right)=y^2\text{Sin}\left(2x\right)$ at target $P\left(\frac{\pi }{4},2\right)$ in the direction of $u=5i+12j.$

For the following exercises, find the directional derivative of the social occasion at point $P$ in the direction of $u$ or $v$ as fit.

263 .

$f\left(x,y\right)=xy,$ $P\left(0,\mathrm{-2}\right),$ $\text{v}=\frac{1}{2}i+\frac{\sqrt{3}}{2}j$

264 .

$h\left(x,y\right)=e^x\text{sin}y,P\left(1,\frac{\pi }{2}\right),v=\text{−}i$

265 .

$h\left(x,y,z\right)=xyz,P\left(2,1,1\right),\text{v}=2i+j-k$

266 .

$f(x,y)=xy,P(1,1),u=\langle \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\rangle$

267 .

$f(x,y)=x^2-y^2,\begin{array}{cc}u=\langle \frac{\sqrt{3}}{2},\frac{1}{2}\rangle ,\hfill & P(1,0)\hfill \end{array}$

268 .

$f(x,y)=3x+4y+7,\begin{array}{cc}u=\langle \frac{3}{5},\frac{4}{5}\rangle ,\hfill & P\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

269 .

$\begin{array}{ccc}f(x,y)=e^x\text{cos}y,\hfill & u=\langle 0,1\rangle ,\hfill & P=\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

270 .

$\begin{array}{ccc}f(x,y)=y^{10},\hfill & u=\langle 0,\mathrm{-1}\rangle ,\hfill & P=(1,\mathrm{-1})\hfill \end{array}$

271 .

$f(x,y)=\text{ln}(x^2+y^2),\begin{array}{cc}u=\langle \frac{3}{5},\frac{4}{5}\rangle ,\hfill & P\left(1,2\right)\hfill \end{array}$

272 .

$f(x,y)=x^{2}y,\begin{array}{cc}P(\mathrm{-5},5),\hfill & v=3i-4j\hfill \end{array}$

273 .

$f(x,y,z)=y^2+xz,\begin{array}{cc}P(1,2,2),\hfill & v=\langle 2,\mathrm{-1},2\rangle \hfill \end{array}$

For the following exercises, find the directional derivative of the subroutine in the direction of the whole vector $u=\text{cos}\theta i+\text{sin}\theta j.$

274 .

$f\left(x,y\right)=x^2+2y^2,\theta =\frac{\pi }{6}$

275 .

$f\left(x,y\right)=\frac{y}{x+2y},\theta =-\frac{\pi }{4}$

276 .

$f\left(x,y\right)=\text{cos}\left(3x+y\right),\theta =\frac{\pi }{4}$

277 .

$w\left(x,y\right)=y{e}^x,\theta =\frac{\pi }{3}$

278 .

$\begin{array}{cc}f\left(x,y\right)=x\text{arctan}(y),\hfill & \theta =\frac{\pi }{2}\hfill \end{array}$

279 .

$\begin{array}{cc}f\left(x,y\right)=\text{ln}(x+2y),\hfill & \theta =\frac{\pi }{3}\hfill \end{array}$

For the following exercises, bump the gradient.

280 .

Discovery the gradient of $f(x,y)=\frac{14-x^2-y^2}{3}.$ Then, find the gradient at point $P\left(1,2\right).$

281 .

Find the gradient of $f(x,y,z)=xy+yz+xz$ at point $P\left(1,2,3\right).$

282 .

Find the gradient of $f(x,y,z)$ at $P$ and in the direction of $u\text{:}$ $f(x,y,z)=\text{ln}(x^2+2y^2+3z^2),\begin{array}{cc}P(2,1,4),\hfill & u=\frac{\mathrm{-3}}{13}i-\frac{4}{13}j-\frac{12}{13}k\hfill \end{array}.$

283 .

$f(x,y,z)=4x^5{y}^2{z}^3,\begin{array}{cc}P(2,\mathrm{-1},1),\hfill & u=\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\hfill \end{array}$

For the following exercises, witness the directional first derivative of the function at show $P$ in the counseling of $Q.$

284 .

$f(x,y)=x^2+3y^2,\begin{array}{cc}P(1,1),\hfill & Q(4,5)\hfill \end{array}$

285 .

$f(x,y,z)=\frac{y}{x+z},\begin{array}{cc}P(2,1,\mathrm{-1}),\hfill & Q(\mathrm{-1},2,0)\hfill \end{array}$

For the chase exercises, find the derivative of the function at $P$ in the direction of $u.$

286 .

$f(x,y)=\mathrm{-7}x+2y,\begin{array}{cc}P(2,\mathrm{-4}),\hfill & u=4i-3j\hfill \end{array}$

287 .

$f(x,y)=\text{ln}(5x+4y),\begin{array}{cc}P(3,9),\hfill & u=6i+8j\hfill \end{array}$

288 .

[T] Use technology to study the level curve of $f(x,y)=4x-2y+3$ that passes through $P(1,2)$ and draw the gradient transmitter at $P.$

289 .

[T] Use engineering to chalk out the grade curve of $f(x,y)=x^2+4y^2$ that passes through $P(\mathrm{-2},0)$ and draw the gradient vector at $P.$

For the following exercises, find the gradient vector at the indicated bespeak.

290 .

$f(x,y)=x{y}^2-y{x}^2,P(\mathrm{-1},1)$

291 .

$f(x,y)=x{e}^y-\text{ln}(x),P\left(\mathrm{-3},0\right)$

292 .

$f(x,y,z)=xy-\text{ln}(z),P(2,\mathrm{-2},2)$

293 .

$f(x,y,z)=x\sqrt{y^2+z^2},P(\mathrm{-2},\mathrm{-1},\mathrm{-1})$

For the following exercises, find the differential coefficient of the role.

294 .

$f(x,y)=x^2+xy+y^2$ at point $\left(\mathrm{-5},\mathrm{-4}\right)$ in the direction the role increases most rapidly

295 .

$f(x,y)=e^{xy}$ at indicate $\left(6,7\right)$ in the direction the part increases virtually rapidly

296 .

$f(x,y)=\text{arctan}\left(\frac{y}{x}\right)$ at stage $\left(\mathrm{-9},9\right)$ in the direction the social function increases nearly rapidly

297 .

$f(x,y,z)=\text{ln}(xy+yz+zx)$ at point $\left(\mathrm{-9},\mathrm{-18},\mathrm{-27}\right)$ in the direction the function increases most speedily

298 .

$f(x,y,z)=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ at pointedness $\left(5,\mathrm{-5},5\right)$ in the direction the function increases most rapidly

For the following exercises, find the maximum rate of change of $f$ at the acknowledged point and the direction in which it occurs.

299 .

$f(x,y)=x{e}^{\text{−}y},$ $\left(1,0\right)$

300 .

$f(x,y)=\sqrt{x^2+2y},$ $\left(4,10\right)$

301 .

$f(x,y)=\text{cos}(3x+2y),\left(\frac{\pi }{6},-\frac{\pi }{8}\right)$

For the undermentioned exercises, find equations of

1. the tangent plane and
2. the normal line to the presumption surface at the given point.

302 .

The even surface $f(x,y,z)=12$ for $f(x,y,z)=4x^2-2y^2+z^2$ at point $\left(2,2,2\right).$

303 .

$f(x,y,z)=xy+yz+xz=3$ at point $\left(1,1,1\right)$

304 .

$f(x,y,z)=xyz=6$ at point $\left(1,2,3\right)$

305 .

$f(x,y,z)=x{e}^y\text{cos lettuce}z-z=1$ at point $\left(1,0,0\right)$

For the next exercises, solve the problem.

306 .

The temperature $T$ in a metal sphere is inversely progressive to the distance from the center of the sphere (the origin: $\left(0,0,0\right)).$ The temperature at point $\left(1,2,2\right)$ is $120\text{°}\text{C}.$

1. Find the plac of change of the temperature at taper off $\left(1,2,2\right)$ in the direction toward orient $\left(2,1,3\right).$
2. Show that, at any point in the celestial sphere, the direction of greatest increase in temperature is given aside a vector that points toward the blood.

307 .

The electric expected (potential) in a certain neighborhood of space is given by the function $V(x,y,z)=5x^2-3xy+xyz.$

1. Feel the rate of deepen of the voltage at item $\left(3,4,5\right)$ in the centering of the vector $\langle 1,1,\mathrm{-1}\rangle .$
2. In which way does the voltage change most rapidly at point $\left(3,4,5\right)?$
3. What is the maximum rate of change of the voltage at point $\left(3,4,5\right)?$

308 .

If the electric potential difference at a point $\left(x,y\right)$ in the XY-plane is $V(x,y)=e^{\mathrm{-2}x}\text{cos}(2y),$ then the galvanic chroma vector at $\left(x,y\right)$ is $E=\text{−}\nabla V(x,y).$

1. Get the electric strength vector at $\left(\frac{\pi }{4},0\right).$
2. Show that, at all pointedness in the plane, the electric likely decreases most rapidly in the direction of the vector $E.$

309 .

In cardinal dimensions, the motion of an saint fluid is governed by a speed potential $\phi .$ The velocity components of the liquid $u$ in the x-direction and $v$ in the y-direction, are minded by $\langle u,v\rangle =\nabla \phi .$ Detect the speed components associated with the velocity potential $\phi (x,y)=\text{hell}\pi x\text{goof}2\pi y.$

find the directional derivative of f x y z

Source: https://openstax.org/books/calculus-volume-3/pages/4-6-directional-derivatives-and-the-gradient